Learn on PengiReveal Math, AcceleratedUnit 12: Area, Surface Area, and Volume

Lesson 12-8: Solve Problems Involving Volume of Spheres

In this Grade 7 Reveal Math Accelerated lesson from Unit 12, students learn to calculate the volume of spheres using the formula V = 4/3πr³ and apply it to real-world problems. Students practice finding a sphere's radius from its circumference before computing volume, and compare volumes across different-sized spheres. The lesson builds on prior knowledge of cone volume to derive the sphere volume formula and reinforces unit tracking throughout multi-step calculations.

Section 1

Volume of a Sphere

Property

The volume of a sphere is given by

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

where rr is the radius of the sphere. Recall that r3r^3, which we read as 'rr cubed,' means r×r×rr \times r \times r.

Examples

  • A gumball has a radius of 1 centimeter. Its volume is V=43π(1)3=43π4.19V = \frac{4}{3} \pi (1)^3 = \frac{4}{3}\pi \approx 4.19 cubic centimeters.
  • A soccer ball has a diameter of 22 cm, so its radius is 11 cm. Its volume is V=43π(11)3=43π(1331)5575.28V = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi (1331) \approx 5575.28 cubic centimeters.
  • A spherical ornament has a volume of 36π36\pi cubic inches. To find its radius, solve 36π=43πr336\pi = \frac{4}{3}\pi r^3, which simplifies to 27=r327 = r^3, so the radius is r=273=3r = \sqrt[3]{27} = 3 inches.

Explanation

Volume measures the space inside a 3D shape, like a ball or a planet. For a sphere, you cube the radius (multiply it by itself three times), then multiply by pi (π\pi), and finally multiply by the fraction 43\frac{4}{3}.

Section 2

Volume of Hemispheres

Property

The volume of a hemisphere is half the volume of a complete sphere:

Vhemisphere=1243πr3=23πr3V_{hemisphere} = \frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3

Examples

Section 3

Volume of a Sphere and Cone

Property

The volume of a sphere is exactly twice the volume of a cone that has the same radius (rr) and a height (hh) equal to the sphere's diameter (2r2r).

Vsphere=2Vconewhen h=2rV_{\text{sphere}} = 2 \cdot V_{\text{cone}} \quad \text{when } h = 2r

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Unit 12: Area, Surface Area, and Volume

  1. Lesson 1

    Lesson 12-1: Describe Cross Sections of Three-Dimensional Figures

  2. Lesson 2

    Lesson 12-2: Understand and Use Square Roots

  3. Lesson 3

    Lesson 12-3: Solve Problems Involving Area and Surface Area

  4. Lesson 4

    Lesson 12-4: Solve Problems Involving Circumference of Circles

  5. Lesson 5

    Lesson 12-5: Solve Problems Involving Area of Circles

  6. Lesson 6

    Lesson 12-6: Understand and Use Cube Roots

  7. Lesson 7

    Lesson 12-7: Solve Problems Involving Volume of Cylinders and Cones

  8. Lesson 8Current

    Lesson 12-8: Solve Problems Involving Volume of Spheres

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Volume of a Sphere

Property

The volume of a sphere is given by

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

where rr is the radius of the sphere. Recall that r3r^3, which we read as 'rr cubed,' means r×r×rr \times r \times r.

Examples

  • A gumball has a radius of 1 centimeter. Its volume is V=43π(1)3=43π4.19V = \frac{4}{3} \pi (1)^3 = \frac{4}{3}\pi \approx 4.19 cubic centimeters.
  • A soccer ball has a diameter of 22 cm, so its radius is 11 cm. Its volume is V=43π(11)3=43π(1331)5575.28V = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi (1331) \approx 5575.28 cubic centimeters.
  • A spherical ornament has a volume of 36π36\pi cubic inches. To find its radius, solve 36π=43πr336\pi = \frac{4}{3}\pi r^3, which simplifies to 27=r327 = r^3, so the radius is r=273=3r = \sqrt[3]{27} = 3 inches.

Explanation

Volume measures the space inside a 3D shape, like a ball or a planet. For a sphere, you cube the radius (multiply it by itself three times), then multiply by pi (π\pi), and finally multiply by the fraction 43\frac{4}{3}.

Section 2

Volume of Hemispheres

Property

The volume of a hemisphere is half the volume of a complete sphere:

Vhemisphere=1243πr3=23πr3V_{hemisphere} = \frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3

Examples

Section 3

Volume of a Sphere and Cone

Property

The volume of a sphere is exactly twice the volume of a cone that has the same radius (rr) and a height (hh) equal to the sphere's diameter (2r2r).

Vsphere=2Vconewhen h=2rV_{\text{sphere}} = 2 \cdot V_{\text{cone}} \quad \text{when } h = 2r

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Unit 12: Area, Surface Area, and Volume

  1. Lesson 1

    Lesson 12-1: Describe Cross Sections of Three-Dimensional Figures

  2. Lesson 2

    Lesson 12-2: Understand and Use Square Roots

  3. Lesson 3

    Lesson 12-3: Solve Problems Involving Area and Surface Area

  4. Lesson 4

    Lesson 12-4: Solve Problems Involving Circumference of Circles

  5. Lesson 5

    Lesson 12-5: Solve Problems Involving Area of Circles

  6. Lesson 6

    Lesson 12-6: Understand and Use Cube Roots

  7. Lesson 7

    Lesson 12-7: Solve Problems Involving Volume of Cylinders and Cones

  8. Lesson 8Current

    Lesson 12-8: Solve Problems Involving Volume of Spheres